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Boeing and the Marshall Space Flight Centre brought to pass many studies of improved Saturn V launch vehicles. The rockets proposed under the MLV and ELV projects would feature longer stages for more propellant capacity, and would be fitted with new, uprated engines.

https://archive.org/stream/NASA_NTRS_Archive_19650020081#page/n13/mode/2up

http://www.astronautix.com/h/hg-3.html

The HG-3 was one such engine. Some basic information on the engine is as follows.

  • Total length = 3.38 metres
  • Nozzle exit diameter = 2.03 metres
  • Vacuum Isp = 451 seconds
  • Vacuum thrust = 314,900lbf (to be increased to 375klbf and 400klbf)
  • Sea level Isp = 280 seconds
  • Chamber pressure = 3,000psia
  • LOX:fuel mixture ratio = 5.1:1 to 5.3:1
  • Operating cycle = Staged combustion

Unfortunately, there is very little other information available on the engine, as it was never built – no test data or proper geometry. However, there are certain traits it shares with other engines that may permit close estimates. The HG-3 ‘formed the basis’ for the design of the SSME, that engine also being a hydrogen/oxygen staged combustion unit with a vacuum Isp of 453 seconds and a very similar chamber pressure. The most powerful 400klbf HG-3s would have been roughly 80,000 pounds short on vacuum thrust.

Interestingly, the HG-3 designs show a very short nozzle that expands quickly. The engine also features a much larger power head and turbopump assembly than the J-2S. enter image description here

So, given this similar performance and close heritage to the RS-25:

  • Could the nozzle area ratio of the HG-3 have been similar to that of the SSME’s 69:1?
  • What causes the large disparity of sea level specific impulse (280s vs 366s) even with the nearly identical chamber pressures?

A sea level-optimised variant is also featured on Astronautix. Vacuum Isp barely changes while sea level Isp rises dramatically to 360 seconds. The only change listed is a lower expansion ratio nozzle.

http://www.astronautix.com/h/hg-3-sl.html

  • Why is the vacuum Isp change so small while the sea level change is so large? (Changing area ratio often does this, but here it seems disproportionally large)

Al.

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  • $\begingroup$ The original SSME design area ratio was 77.5. It dropped when they went to the Large Throat Main Combustion Chamber (first flight 2001). $\endgroup$ – Organic Marble Jan 6 '17 at 14:01
  • $\begingroup$ @OrganicMarble Ah yes, thought that figure was familiar. Do you think that the HG-3 may have sported the original combustion chamber dimensions of the SSME? I've read that it was developed from the HG-3, but I'm not sure how close that relationship was. Perhaps the larger RS-25 chamber was to assist in achieving thrust levels above 500,000 pounds, well beyond the power of its forerunner. $\endgroup$ – Alastair Haslam Jan 6 '17 at 14:56
  • $\begingroup$ I had not heard of the HG-3 until this post, I will have to read up on it. $\endgroup$ – Organic Marble Jan 6 '17 at 14:57
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Here is a mathematical answer to your question. Using the numbers you provided for the HG-3, we have,

$$I_{sp} = 451 \ \text{s} \ (\text{vac})$$ $$F = 1,400,745 \ \text{N} \ \ (314,900 \ \text{lbf}) \ (\text{vac})$$

From this alone, we can get an estimate of the steady mass flow rate through the engine,

$$ \dot{m} = \frac{F}{I_{sp} g} = 317 \ \text{kg/s}$$

Now we can figure out the size of the throat from the choked flow condition at the nozzle throat. The mass flow parameter can be rearranged to the form,

$$ A_t = \frac{\dot{m} \sqrt{T_1}} {p_1} \sqrt{\frac{R}{\gamma}} \left(\frac{\gamma+1}{2}\right)^{\frac{(\gamma+1)}{2(\gamma-1)}}$$

where $\dot{m}$ is the flow rate through the engine, $T_1$ is the chamber temperature, $p_1$ is the chamber pressure, $R$ is the specific gas constant of the burned products, and $\gamma$ is the ratio of specific heats of the burned products. Using the NASA Chemical Equilibrium with Applications (CEA) code for an average LOX:fuel mixture ratio of 5.2:1 at a chamber pressure of 3,000 psia, we obtain,

$$ T_1 = 3,414 \ \text{K} $$ $$ R = 678 \ \text{J/kg$\cdot$K} $$ $$ \gamma = 1.16 $$

Substituting these values with the previously calculated mass flow rate yields,

$$ A_t = A^{*} = 0.0363 \ \text{m$^2$} \ \ (0.3910 \ \text{ft$^2$})$$ $$ d_t = d^{*} = 0.215 \ \text{m} \ \ (0.705 \ \text{ft})$$

Now from your provided information, we were given that $d_e$ = 2.03 m, namely, $A_e$ = 3.235 m$^2$. So our nozzle expansion area ratio required to obtain a vacuum $I_{sp} = 451 \ \text{s}$ is given by,

$$ \boxed{ \epsilon = \frac{A_e}{A_t} = \frac{A_e}{A^*} = 89.1 }$$

That is the answer to your first question.

Now we want to look at the sea-level performance. We will assume the mass flow rate through the engine is similar to the vacuum scenario, since the LOX/fuel mixture ratio only fluctuates by $\pm$ 1 and we already took the average for sizing the nozzle throat above. We need to determine the nozzle exit parameters. For idealized quasi one-dimensional compressible flow, the Area-Mach relation can be used to estimate the nozzle exit Mach number.

$$\frac{A_e}{A_t} = \frac{A_e}{A^*} = \frac{1}{M_e}\left[\frac{2}{\gamma+1} \frac{\gamma-1}{2}M_e^2\right]^{\frac{\gamma+1}{2(\gamma-1)}}$$

Using the Newton-Raphson numerical technique to solve the above for $M_e$ given $A_e/A_t$ = 89.1 we obtain,

$$ M_e = 4.52 $$

It is conventional in idealized rocket analysis to assume isentropic expansion of the burned products through the nozzle. Hence, from isentropic relations we have at the nozzle exit plane,

$$ p_e = p_1 \left(1 + \frac{\gamma-1}{2} M_e^2 \right)^{-\frac{\gamma}{\gamma-1}} $$ $$ T_e = T_1 \left(1 + \frac{\gamma-1}{2} M_e^2 \right)^{-1} $$

To which we obtain,

$$p_e = 18,735 \ \text{Pa} \ \ (2.72 \ \text{psia}) $$ $$T_e = 1,285 \ \text{K} \ \ (1,853 \ \text{$^{\circ}$F}) $$

Further, the sound speed on the nozzle exit plane is given by,

$$a_e = \sqrt{\gamma R T_e} $$

To which we obtain,

$$ a_e = 1,006.5 \ \text{m/s} \ \ (3,302 \ \text{ft/s}) $$

Hence, our nozzle exit velocity is simply,

$$ V_e = a_e M_e = 4,551 \ \text{m/s} \ \ (14,931 \ \text{ft/s}) $$

Now our sea-level thrust is given by,

$$ F = \dot{m} V_e + \left(p_e - p_a\right) A_e $$

where at sea-level, $p_a$ = 101,325 Pa (14.7 psia). Substituting all of the calculated parameters yields,

$$ F = 1,173,595 \ \text{N} \ \ (263,834 \ \text{lbf} ) $$

Similarly, the sea-level specific impulse is given by,

$$ I_{sp} = 378 \ \text{s} $$

This is obviously based on idealized rocket performance analysis. Additionally, the sea-level specific impulse is very large compared to the 280 you presented in the question. So what is going on here? Well, it turns out we are going to suffer performance losses from over-expansion with a nozzle that has a nozzle expansion area ratio of roughly 89. According to Sutton (Rocket Propulsion Elements), if $p_a$ is slightly higher than $p_e$ then we will have our idealized $I_{sp} \approx$ 378 and our nozzle will continue to flow full. He further states that the nozzle will continue to flow full while developing a slight contraction until the nozzle exit pressure drops to about 40% of the ambient. However, in our case we have very severe over-expansion at the sea-level condition, where our nozzle exit pressure is only approximately 18% of the sea-level ambient back pressure condition. According to the Summerfield criterion (Flow Separation in Over-expanded Supersonic Exhaust Nozzles”, Jet propulsion, Vol. 24, No. 9, page 319-321, 1954), which is a general rule of thumb, the nozzle flow will separate at only $p_e/p_a$ = 0.4. So in our case, we will definitely have flow separation in the divergent portion of the nozzle, which will drastically undercut our anticipated ideal $I_{sp}$. Below is a schematic from Sutton's text that demonstrates the effects of flow separation in the diverging portion of the nozzle.

enter image description here

The general answer to your questions go as follows:
1.) Yes the HG-3 engine could achieve a vacuum $I_{sp} = 451 \ \text{s}$ provided the nozzle expansion area ratio was roughly $\epsilon = 89$.

2.) Nozzle expansion area ratio plays a large role in the over-expansion effects regarding the nozzle flow losses and separation. Cutting back the expansion area ratio to something similar to the SSME (69) alleviates this issue at the sea-level condition.

3.) The sensitivity of sea-level performance to over-expansion is much more severe than the performance gains from higher expansion area-ratios in a vacuum back pressure condition. From the NASA CEA code and the conditions previously mentioned, the vacuum specific impulse goes as follows:

$$ \begin{array}{c|c} \epsilon = \frac{A_e}{A_t} & I_{sp} - \text{s} \ (\text{vac}) \\ \hline 60 & 445 \\ 70 & 447 \\ 80 & 449 \\ 90 & 451 \\ 100 & 452 \end{array} $$

I could also tabulate the sea-level performance, however, it is also the ideal sea-level performance and won't account for the performance losses due to over-expansion effects. Determining the losses due to over-expansion effects requires either full numerical solutions to the Navier-Stokes equations, or experimental testing.

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  • $\begingroup$ this answer is beyond outstanding! You have solved the issues mathematically and have also explained every step of the way to make it easy to grasp. Very helpful, thank you! $\endgroup$ – Alastair Haslam Jan 8 '17 at 4:14
  • $\begingroup$ With further reading, I may have to run some new values through to you. It seems the true mixture ratio intended for the HG-3 was actually lower at 5.0:1 LOX:LH2. How greatly would this alter the optimum nozzle area ratio to achieve 451s vacuum Isp? What about the change to combustion chamber temperature? $\endgroup$ – Alastair Haslam Jan 8 '17 at 4:19
  • $\begingroup$ @AlastairHaslam Changing the LOX/fuel mixture ratio to 5.0 with a chamber pressure of $p_1$ = 3,000 psia yields the following: $T_1$ = 3,354 K, $R$ = 698.5 J/kg$\cdot$K, and $\gamma$ = 1.17. Using these new numbers, the new optimum nozzle expansion area ratio for the vacuum specific impulse of 451 s is $\epsilon$ = 88.6. Notice, the solution doesn't change all that much, as dropping the LOX/fuel mixture ratio from 5.2 to 5.0 only drops the chamber temperature by approximately 60 K. $\endgroup$ – TRF Jan 8 '17 at 4:49
  • $\begingroup$ Interesting to see a near-identical nozzle expansion ratio. But it surprises me to see the chamber temperature is a little lower than I would have thought. For instance, the standard J-2 at only a slightly higher 5.5:1 LOX:fuel mix and a much lower 763psia chamber pressure operates at 3,450K. Is this just because there are always subtle differences between the calculated optimum and real-life testing? $\endgroup$ – Alastair Haslam Jan 8 '17 at 11:07
  • $\begingroup$ (Sorry for all the questions, but) assuming the same nozzle exit area and engine dimensions, what would change if the engine were uprated to 375klbf thrust? I assume chamber pressure would increase, and the HG-3 would possibly require a larger engine throat area similar to the upgrades on late-model SSMEs. $\endgroup$ – Alastair Haslam Jan 8 '17 at 11:11

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